We will proceed to solve each case to determine the solution of the problem.
case a) \(x^{2}+2x+4=0\)
Group terms that contain the same variable, and move the constant to the opposite side of the equation
\(x^{2}+2x=-4\)
Complete the square. Remember to balance the equation by adding the same constants to each side.
\(x^{2}+2x+1=-4+1\)
\(x^{2}+2x+1=-3\)
Rewrite as perfect squares
\((x+1)^{2}=-3\)
\((x+1)=(+/-)\sqrt{-3}\\(x+1)=(+/-)\sqrt{3}i\\x=-1(+/-)\sqrt{3}i\)
therefore
case a) is not the solution of the problem
case b) \(x^{2}-2x+4=0\)
Group terms that contain the same variable, and move the constant to the opposite side of the equation
\(x^{2}-2x=-4\)
Complete the square. Remember to balance the equation by adding the same constants to each side.
\(x^{2}-2x+1=-4+1\)
\(x^{2}-2x+1=-3\)
Rewrite as perfect squares
\((x-1)^{2}=-3\)
\((x-1)=(+/-)\sqrt{-3}\\(x-1)=(+/-)\sqrt{3}i\\x=1(+/-)\sqrt{3}i\)
therefore
case b) is not the solution of the problem
case c) \(x^{2}+2x-4=0\)
Group terms that contain the same variable, and move the constant to the opposite side of the equation
\(x^{2}+2x=4\)
Complete the square. Remember to balance the equation by adding the same constants to each side.
\(x^{2}+2x+1=4+1\)
\(x^{2}+2x+1=5\)
Rewrite as perfect squares
\((x+1)^{2}=5\)
\((x+1)=(+/-)\sqrt{5}\\x=-1(+/-)\sqrt{5}\)
therefore
case c) is not the solution of the problem
case d) \(x^{2}-2x-4=0\)
Group terms that contain the same variable, and move the constant to the opposite side of the equation
\(x^{2}-2x=4\)
Complete the square. Remember to balance the equation by adding the same constants to each side.
\(x^{2}-2x+1=4+1\)
\(x^{2}-2x+1=5\)
Rewrite as perfect squares
\((x-1)^{2}=5\)
\((x-1)=(+/-)\sqrt{5}\\x=1(+/-)\sqrt{5}\)
therefore
case d) is the solution of the problem
therefore
the answer is
\(x^{2}-2x-4=0\)
